91Ó°ÊÓ

The rate of A random variable following the Poisson process is 5 events per hour.

We know that the proofs of the results in the Poison process have established that the models the time to wait for the first event. It is because of the memoryless property of this distribution.

Therefore, the waiting time T₁until the first event occurs follows an exponential distribution with the same parameter as the Poisson process.

Therefore,

\({T_1} \sim \exp \left( {\beta = 5} \right)\)

We know that the proofs of the results in the Poison process have established that the Gamma distribution is the sum of some Exponential distributions and thus is the waiting time distribution forkevents.

Now,

\({T_1} \sim \exp \left( {\beta = 5} \right)\)

And we know that

\(Gamma\left( {\alpha ,\beta = 5} \right) = \sum\limits_\alpha {Exp\left( {\beta = 5} \right)} \)

Since we are waiting for the first k events, \({T_k}\), \(\alpha = k\).

Therefore, the distribution follows Gamma distribution, that is

\({T_k} \sim Gamma\left( {\alpha = k,\beta = 5} \right)\)

20 minutes are to be converted to hours:

\(\begin{array}{c}t = \frac{{20}}{{60}}\\ = \frac{1}{3}\end{array}\)

The parameter value is 5, with the occurrence of first k events not happening.

The parameter value hence changes to k-1.

Therefore, the distribution is,

\(\begin{array}{l} = {e^{ - \lambda t}}\\ = {e^{ - \frac{{5\left( {k - 1} \right)}}{3}}}\end{array}\)